Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Asked by Abhisek | 1 year ago |  125

##### Solution :-

Let us consider the equation of the required circle be

(x – h)+ (y – k)2 = r2

We know that the circle passes through points (2,3) and (-1,1).

(2 – h)2+ (3 – k)2 =r2 ……………..(1)

(-1 – h)2+ (1– k)2 =r2 ………………(2)

Since, the centre (h, k) of the circle lies on line x – 3y – 11= 0,

h – 3k =11………………… (3)

From the equation (1) and (2), we obtain

(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2

4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2

4 – 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11……………. (4)

Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,

6h+ 4k – 6(h-3k) = 11 – 66

6h + 4k – 6h + 18k = 11 – 66

22 k = – 55

K = $$\dfrac{-5}{2}$$

Substitute this value of K in equation (4) to get,

6h + 4($$\dfrac{-5}{2}$$) = 11

6h – 10 = 11

6h = 21

h = $$\dfrac{21}{6}$$

h = $$\dfrac{7}{2}$$

We obtain h = $$\dfrac{7}{2}$$and k =$$\dfrac{-5}{2}$$

On substituting the values of h and k in equation (1), we get

(2 – $$\dfrac{7}{2}$$)2 + (3 + $$\dfrac{5}{2}$$)2 = r2

$$(\dfrac{4-7}{2})^2+(\dfrac{6+5}{2})^2=r^2$$

$$(\dfrac{-3}{2})^2$$ + $$(\dfrac{11}{2})^2$$= r2

$$\dfrac{9}{4}$$ + $$\dfrac{121}{4}$$ = r2

$$\dfrac{130}{4}$$ = r2

The equation of the required circle is

(x – $$\dfrac{7}{2}$$)2 + (y + $$\dfrac{5}{2}$$)2 = $$\dfrac{130}{4}$$

$$(\dfrac{2x-7}{2})^2+(\dfrac{2y+5}{2})^2=\dfrac{130}{4}$$

4x2 -28x + 49 +4y2 + 20y + 25 =130

4x2 +4y2 -28x + 20y – 56 = 0

4(x2 +y2 -7x + 5y – 14) = 0

x+ y– 7x + 5y – 14 = 0

The equation of the required circle is x+ y– 7x + 5y – 14 = 0

Answered by Pragya Singh | 1 year ago

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